MATH 132 EXAM II ( FALL 2010Solutions)1) Evaluate '-9= ÐBÑ.BÞ$EÑ -9= ÐBÑ ! -9= ÐBÑ ! GFÑ-9=ÐBÑ! -9= ÐBÑ ! G#% $""%$ GÑ B " -9=ÐBÑ ! GHÑ=38ÐBÑ" =38 ÐBÑ ! GIÑ=38ÐBÑ" =38 ÐBÑ ! G"" " "## % $#% $ JÑ B ! =38ÐBÑ ! GKÑ -9=ÐBÑ! GLÑ-9=ÐBÑ" -9=ÐBÑ =38 ÐBÑ ! G"" "## %%# MÑ =38ÐBÑ " =38ÐBÑ -9= ÐBÑ ! GNÑ=38ÐBÑ" -9= ÐBÑ ! G##""#$ solution: = ''-9= ÐBÑ -9=ÐBÑ .B œ Ð" " =38 ÐBÑÑÐ-9=ÐBÑÑ .B Ð ? œ =38ÐBÑ##.? œ -9=ÐBÑ. B Ñ œ " " ?.?œ?" ?œ=38ÐBÑ" =38 ÐBÑ ! GÐIÑ'#$ $""$$--------------------------------------------------------------------------------------------2) What is the form of the integral after we make the trigonometric'B""B##È.Bß substitution, Bœ=38Ð ÑÞ)EÑ =38Ð Ñ. FÑ -9=Ð Ñ. GÑ =38 Ð Ñ. HÑ -9= Ð Ñ. IÑ =38Ð Ñ-9=Ð Ñ.''''')) )) )) )) ) )) ##JÑ =38 Ð Ñ-9=Ð Ñ. KÑ =38Ð Ñ-9= Ð Ñ. LÑ >+8Ð Ñ. MÑ =/- Ð Ñ.''''## #) )) ) )) )) )) JÑ=/-ÐÑ>+8ÐÑ.')))solution: = ''=38 Ð Ñ""=38 Ð Ñ###))È-9=Ð Ñ. œ =38 Ð Ñ . ! GÐGÑ)) ) )-------------------------------------------------------------------------------------------------------$Ñ .B Ð B # !ÑÞUse to evaluate partial fractions'B ! #B ! B#EÑ 68Ð#B ! "Ñ FÑ 68ÐB ! BÑ GÑ +<->+8ÐBÑ HÑ IÑ Ð#B ! "Ñ ##"+<->+8ÐBÑJ Ñ 68Ð Ñ KÑ 68Ð Ñ LÑ 68 Ð Ñ MÑ 68Ð+<->+8ÐBÑÑ N Ñ +<- >+8Ð6+8ÐBÑÑB!"B #BBB!"B"" #solution: = '#" BBB!"B!"" .B œ #68ÐBÑ " 68ÐB ! "Ñ œ 68Ð Ñ ! GÐKÑ#--------------------------------------------------------------------------------------------------4) Approximate using the Trapezoidal Rule with to 4 decimal places'"#"B.Bß 8 œ %ß ÞEÑ !Þ'* ! FÑ !Þ'*'" GÑ !Þ'*&$ HÑ !Þ'*%! IÑ !Þ'*$# J Ñ !Þ'*##5KÑ !Þ'*"( LÑ !Þ'*"" MÑ !Þ'*(! N Ñ !Þ'*!! Ð B œ ß œ Ñ?#""B"%#)?solution: = ("""""") " &Î% $Î# (Î% #! #Ð Ñ ! #Ð Ñ ! #Ð Ñ ! Ñ œ !Þ'*(!#$)!*# ÐMÑ2.5) How large should n be to guarantee that Simpson's Rule of approximation for is acurate to within 0.0001 ? ( recall |E'!#"#BOÐ,"+Ñ")! 8/.B À lŸ ÑW&%EÑ( FÑ) GÑ* HÑ"! IÑ"" JÑ"# KÑ"$ LÑ"% MÑ"& NÑ"'solution: We want 0 ÐBÑ œ "'/ Ÿ "' 98 Ò!ß #Ó Ê O œ "'Þ Ÿ !Þ!!!"ÞÐ%Ñ "#B"'Ð# Ñ")! 8&%X2/8 8 $ Ê8$ Ð Ñ œ "#Þ*)'( Ê 8 œ "%Þ ÐLÑ% "Î%"'Ð$#Ñ "'Ð$#Ñ")!Ð!Þ!!!"Ñ ")!Ð!Þ!!!"Ñ----------------------------------------------------------------------------------------------6) Evaluate the improper integral if it is convergent. If not say it is'!#"B"#"B/.Bß divergent.EÑ FÑ GÑ / HÑ / IÑ J Ñ / KÑ " "LÑ " """ " " "/ ///#$Î#/È È#$Î#ÈMÑ / " "NÑ.3@/<1/8>#solution: '+#"B" " "" "" "+Ä!#" " "" "" "B B #+ #+ ##+!/.Bœ/lœ/ " /Þ / " /œ/ÞÐEÑlim------------------------------------------------------------------------------------------------7) Find the area of the region enclosed by the curves CœB +8. Cœ " B ! %B##EÑ FÑ GÑ HÑ IÑ J Ñ KÑ LÑ MÑ N Ñ"#"$%& & )"!"$$$##$# $ $$ $solution: Intersection points are (0,0) and (2,4). , is upper curve, and0ÐBÑ œ " B ! %B# is the lower curve. Then Area1ÐBÑ œ B œ Ð " B ! %BÑ " B.Bœ###!#''!##)$%B " #B .B œ Þ ÐLÑ--------------------------------------------------------------------------------------------------9)Find the volume of the solid obtained by rotating the region bounded by the curve y = 2 and the line y = x .ÈB about the x-axisA) B) 11 1111"$%1 20 2 3 53 3 511 1 1GÑ "# HÑ IÑ #% J Ñ KÑ $# LÑ %! MÑ N Ñ %)solution: The points of intersection are (0,0) and (4,4). Using the disc method,0ÐBÑ œ # B Bß 1ÐBÑ œ BßÈÈ is further from the line of rotation, with distance 2 and is closer, with ditance x. Then the volume is 1'!%#$#$%B " B.Bœ ÞÐJÑ1------------------------------------------------------------------------------------------------------10)Find the volume of the solid obtained by rotating the region bounded by the curve and the line .BœC Bœ" Bœ"#about the lineA) B) D 111 1 1""$$"'"&"&& "1 2 2 5 51111 1GÑ# Ñ IÑ% JÑ KÑ LÑ%! MÑ NÑ%)solution: The points of intersection are (1,1) and (1, The distance from " "ÑÞ 0ÐCÑ œ C#and the line is , so using the disc method we get the volumeBœ" "" C# œÐ"" CÑ .Cœ "" #C ! C.Cœ# "" #C ! C.Cœ ÞÐMÑ11 1'' '"" ""!"" "## #% #%"'$1$Þ""Ñ Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the curve and the lineCœ$! #B " B# B ! Cœ$ Þabout the -axisCA) B) G11 1#( ) $ $!#$&( 2 2 1111GÑ# HÑ IÑ % JÑ ÑLÑ %! MÑ N Ñ %) 11$&% 1solution: For cylindrical shell about the y-axis you must describe the region in terms offunctions of x. is the top curve and is the bottom0ÐBÑ œ $ ! #B " B 1ÐBÑ œ $ " Bß#curve. The intersection points are (0,3) and (3,0). For each x between 0 and 3, thedistance to the y-axis is x. Therefore the volume is given by 211''!!$$##$#(#BÐÐ$ ! #B " BÑ" Ð$ " BÑÑ .B œ # $B " B.Bœ ÞÐFÑ1-------------------------------------------------------------------------------------------------------12) Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the curve and the lines Cœ B Cœ!ß Bœ"ßÈ .about the line Cœ " "A) B) 11 1 1 1""$$"$'& "&&)1 2 2 11 111GÑ # HÑ IÑ % J Ñ KÑ LÑ %! MÑ N Ñ %)solution: About the line , we must use the region between function of y. (1,1) isCœ " "the intersection point of and with , the top, and , theBœ" BœC ß Bœ" BœC##bottom. For each y between 0 and 1, the distance to is Cœ " "C! "ÞZœ# ÐC! "ÑÐ" " CÑ.Cœ# C! " " C " C.Cœ ÞÐFÑ11''!!""#$#""'1----------------------------------------------------------------------------------------------------------13) Which of the following integrals gives the length of the curve ?Bœ>" >ßCœ> ß!Ÿ>Ÿ$# $Î#%$A) '' 'ÈÈÈ!! !$$ $## #$ ! %> .> FÑ " ! %> .> GÑ $ ! &> .>HÑ " ! &> .> IÑ $ ! $> .> J Ñ " ! $> .>'''ÈÈÈ!!!$$$###KÑ $ ! >.>LÑ "! >.>MÑ $! #> .> N Ñ " ! #> .>''' 'ÈÈÈ È!!! !$$$ $## # # solution: .B .B.> .> .> .>## #.C .Cœ"" #>ß Ð Ñ œ " " %> ! %> ß œ #> ß Ð Ñ œ %>Þ"#Pœ Ð Ñ ! ÐÑ.>œ "! %> .>Þ ÐFÑ''ÉÈ!!$$.B.> .>##
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